A 190-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.700 rev/s in 2.00 s? (State the magnitude of the force.)

Answer 1

F*r = I * a

Where F is the force on the merry-go-round, r is the radius, I is the moment of inertia of the merry-go-round, and a is the angular acceleration

for a solid, horizontal disk, the moment of inertia is: 1/2*m*r^2

F*r = 1/2*m*r^2*a

F = 1/2*m*r*a

w = w0 + a*t
where w is the angular speed, t is time, w0 is the initial angular speed

0.7 rev / s *(2*pi radians/rev) = 4.3 radians/s

4.3 = a*2

a = 2.2 rad/s^2

Plug this into the above equation to solve for F

Answer 2

wrong